# $K\in \mathbb{R}^{n}$ is compact, show that $K \subset K(r,\delta)$

Consider $$\mathbb{R}^{n}$$ with natural topology. Let $$K\in \mathbb{R}^{n}$$ compact set and $$K \subset \bigcup _{n\in \mathbb{N}}K(r_{n},1)$$, where $$K(r_{n},1)$$ is open ball (in Euclidean metric). Show that exists $$r\in \mathbb{R}^{n}$$ and $$\delta>0$$ such as $$K\subseteq K(r,\delta)$$

My solution is:

So if $$K$$ is compact, then:

1) $$K$$ is Hausdorff space,

2) from every open cover there is a finite subcover.

Now, $$\bigcup _{n\in \mathbb{N}}K(r_{n},1)$$ is finite cover of $$K$$. As we know, $$K$$ is compact, so there exists finite subcover of $$\bigcup _{n\in \mathbb{N}}K(r_{n},1)$$, $$r\in \mathbb{R}^{n}$$ and $$\delta>0: K(r,\delta)$$

We concude that:

$$K\subset \bigcup _{n\in \mathbb{N}}K(r_{n},1) \subset K(r,\delta)$$

Am I correct?

• $K\color{red}{\in}\Bbb R^n$ is incorrect. – Chris Custer Jan 23 at 1:17

You did not say what $$r$$ and $$\delta$$ are. Take $$r=0$$ and $$\delta=1+\max \{\|r_i\|: 1\leq i \leq n\}$$. Note that $$x \in K$$ implies $$\|x-r_i\| <1$$ for some $$i$$. This gives $$\|x-0\| \leq \|r_i\|+\|x-r_i\|<1+\|r_i\| <\delta$$.
• Can I take $r=0$? Such ball does not exist? – Michal Jan 22 at 23:52
• @Michal By $r=0$ I mean $r =(0,0,...,0)$. This vector belongs to $\mathbb R^{n}$ so $K(r,\delta)$ exists for any $\delta >0$. – Kavi Rama Murthy Jan 22 at 23:54
• Forgive me, for a second I thought $r$ was radius... Thanks for clarification. – Michal Jan 22 at 23:57